Rod Cutting Problem in DSA Javascript

Mental Model

We want to cut a rod into pieces to get the highest total price by choosing the best lengths to cut.

Analogy: Imagine you have a chocolate bar and prices for different piece sizes. You want to break it into pieces to get the most money when selling.

Rod length: 8 units
Prices: [1, 5, 8, 9, 10, 17, 17, 20]
Rod: [8 units]
Goal: Cut into pieces to maximize total price

Dry Run Walkthrough

Input: rod length = 4, prices = [1, 5, 8, 9]

Goal: Find the best way to cut the rod of length 4 to get the highest price

Step 1: Calculate max price for rod length 1

maxPrices[1] = 1

Why: Base case: price for length 1 is known

Step 2: Calculate max price for rod length 2 by checking cuts: 1+1 or 2

maxPrices[2] = max(1+1, 5) = 5

Why: Choose the best between cutting into two pieces of length 1 or one piece of length 2

Step 3: Calculate max price for rod length 3 by checking cuts: 1+2, 2+1, or 3

maxPrices[3] = max(1+5, 5+1, 8) = 8

Why: Choose the best combination of cuts or no cut

Step 4: Calculate max price for rod length 4 by checking cuts: 1+3, 2+2, 3+1, or 4

maxPrices[4] = max(1+8, 5+5, 8+1, 9) = 10

Why: Find the best total price by trying all cut combinations

Result:

maxPrices array: [0, 1, 5, 8, 10]
Maximum price for rod length 4 is 10

Annotated Code

DSA Javascript

class RodCutting {
  constructor(prices) {
    this.prices = prices;
  }

  maxProfit(length) {
    const maxPrices = Array(length + 1).fill(0);

    for (let i = 1; i <= length; i++) {
      let maxVal = -Infinity;
      for (let j = 1; j <= i; j++) {
        const currentVal = this.prices[j - 1] + maxPrices[i - j];
        if (currentVal > maxVal) {
          maxVal = currentVal;
        }
      }
      maxPrices[i] = maxVal;
    }

    return maxPrices[length];
  }
}

// Driver code
const prices = [1, 5, 8, 9];
const rodLength = 4;
const rodCutting = new RodCutting(prices);
console.log(rodCutting.maxProfit(rodLength));

for (let i = 1; i <= length; i++) {

Iterate over all rod lengths from 1 to target length

for (let j = 1; j <= i; j++) {

Try all possible first cut lengths for current rod length i

const currentVal = this.prices[j - 1] + maxPrices[i - j];

Calculate price for first cut plus best price for remaining length

if (currentVal > maxVal) { maxVal = currentVal; }

Update max price if current cut gives better profit

maxPrices[i] = maxVal;

Store best price found for rod length i

Output

10

Complexity Analysis

Time: O(n^2) because for each length i, we try all cuts j from 1 to i

Space: O(n) for storing max prices for each rod length up to n

vs Alternative: Naive recursive approach has exponential time due to repeated calculations; dynamic programming stores results to avoid repeats

Edge Cases

rod length 0

Returns 0 because no rod means no profit

DSA Javascript

const maxPrices = Array(length + 1).fill(0);

rod length 1

Returns price for length 1 directly from prices array

DSA Javascript

maxPrices[i] = maxVal;

prices array shorter than rod length

Only prices for available lengths are used; no error occurs

DSA Javascript

const currentVal = this.prices[j - 1] + maxPrices[i - j];

When to Use This Pattern

When you need to find the best way to split something into parts to maximize or minimize a value, use dynamic programming to try all splits efficiently.

Common Mistakes

Mistake: Not storing intermediate results and recalculating subproblems repeatedly
Fix: Use an array to store max prices for all lengths up to n to avoid repeated work

Mistake: Off-by-one errors when accessing prices array
Fix: Remember prices index is j-1 when cutting length j

Summary

Find the best way to cut a rod to get the highest total price.

Use when you want to maximize profit by splitting a resource into parts.

Try all cuts and store best results to avoid repeated calculations.

Practice

1. What is the main goal of the Rod Cutting Problem in algorithms?

easy

A. To count the total number of ways to cut the rod

B. To minimize the number of cuts made on the rod

C. To find the shortest rod length possible

D. To maximize the total profit by cutting the rod into pieces

Solution

1. Step 1: Understand the problem objective

   The rod cutting problem aims to cut a rod into pieces to get the highest total price from selling those pieces.

2. Step 2: Identify the goal

   The goal is to maximize profit, not minimize cuts or count ways.

3. Final Answer:

   To maximize the total profit by cutting the rod into pieces -> Option D

4. Quick Check:

   Maximize profit = To maximize the total profit by cutting the rod into pieces [OK]

Hint: Remember: maximize profit, not minimize cuts [OK]

Common Mistakes:

• Thinking the goal is to minimize cuts
• Confusing number of ways with maximum profit
• Assuming rod length is to be minimized

2. Which of the following is the correct way to declare a function in JavaScript to solve the rod cutting problem?

easy

A. function rodCutting(prices, n) { /* code */ }

B. def rodCutting(prices, n): /* code */

C. func rodCutting(prices, n) { /* code */ }

D. rodCutting function(prices, n) { /* code */ }

Solution

1. Step 1: Identify JavaScript function syntax

   JavaScript functions are declared using the function keyword followed by the name and parameters.

2. Step 2: Check each option

   function rodCutting(prices, n) { /* code */ } uses correct JavaScript syntax. def rodCutting(prices, n): /* code */ is Python, C is Go/Swift style, D is invalid.

3. Final Answer:

   function rodCutting(prices, n) { /* code */ } -> Option A

4. Quick Check:

   JS function syntax = function rodCutting(prices, n) { /* code */ } [OK]

Hint: JS functions start with 'function' keyword [OK]

Common Mistakes:

• Using Python or other language syntax
• Missing 'function' keyword
• Incorrect parameter declaration

3. What will be the output of the following code snippet?

const prices = [1, 5, 8, 9];
function rodCutting(prices, n) {
  if (n === 0) return 0;
  let maxVal = -Infinity;
  for (let i = 1; i <= n; i++) {
    maxVal = Math.max(maxVal, prices[i - 1] + rodCutting(prices, n - i));
  }
  return maxVal;
}
console.log(rodCutting(prices, 4));

medium

A. 8

B. 9

C. 10

D. 15

Solution

1. Step 1: Understand the recursive calls for n=4

   Calculate max profit for rod length 4 by checking all cuts:
   - Cut 1: price[0] + rodCutting(3) = 1 + 8 = 9
   - Cut 2: price[1] + rodCutting(2) = 5 + 5 = 10
   - Cut 3: price[2] + rodCutting(1) = 8 + 1 = 9
   - Cut 4: price[3] + rodCutting(0) = 9 + 0 = 9

2. Step 2: Calculate rodCutting(3), rodCutting(2), rodCutting(1)

   rodCutting(3) = max(1+5=6, 5+1=6, 8+0=8) = 8
   rodCutting(2) = max(1+1=2, 5+0=5) = 5
   rodCutting(1) = 1

3. Final Answer:

   10 -> Option C

4. Quick Check:

   Max profit for length 4 = 10 [OK]

Hint: Check all cuts and pick max sum recursively [OK]

Common Mistakes:

• Ignoring recursive calls
• Using wrong indices for prices
• Assuming max is price of full rod only

4. Identify the error in the following rod cutting function and select the fix:

function rodCutting(prices, n) {
  if (n <= 0) return 0;
  let maxVal = 0;
  for (let i = 1; i < n; i++) {
    maxVal = Math.max(maxVal, prices[i - 1] + rodCutting(prices, n - i));
  }
  return maxVal;
}

medium

A. Change loop condition to i <= n

B. Initialize maxVal to -Infinity

C. Change base case to if (n === 0) return 1

D. Remove recursive call inside loop

Solution

1. Step 1: Analyze loop range

   The loop runs from i=1 to i < n, so it misses the case when i = n (full rod cut).

2. Step 2: Fix loop condition

   Changing to i <= n includes the full rod cut option, ensuring all possibilities are checked.

3. Final Answer:

   Change loop condition to i <= n -> Option A

4. Quick Check:

   Loop must include full length cut = Change loop condition to i <= n [OK]

Hint: Loop must cover full rod length cuts [OK]

Common Mistakes:

• Using i < n instead of i <= n
• Incorrect base case return value
• Removing recursion breaks logic

5. Given a rod of length 5 and prices array [2, 5, 7, 8, 10], what is the maximum profit using dynamic programming approach?

hard

A. 14

B. 17

C. 15

D. 12

Solution

1. Step 1: Use bottom-up DP to calculate max profit

   Calculate max profit for lengths 1 to 5:
   Length 1: max = 2
   Length 2: max = max(5, 2+2) = 5
   Length 3: max = max(7, 5+2, 2+5) = 7
   Length 4: max = max(8, 7+2, 5+5, 2+7) = 10
   Length 5: max = max(10, 8+2, 7+5, 5+7, 2+10) = 17

2. Step 2: Confirm maximum profit for length 5

   The best cut is length 1 + length 4 (2+15=17) or length 2 + length 3 (5+7=12), so 17 is maximum.

3. Final Answer:

   17 -> Option B

4. Quick Check:

   DP max profit for length 5 = 17 [OK]

Hint: Use DP table from length 1 to n for max profit [OK]

Common Mistakes:

• Not considering all cut combinations
• Using recursive without memoization causing slow results
• Confusing prices index with rod length